linprog | R Documentation |
Solves simple linear programming problems, allowing for inequality and equality constraints as well as lower and upper bounds.
linprog(cc, A = NULL, b = NULL, Aeq = NULL, beq = NULL,
lb = NULL, ub = NULL, x0 = NULL, I0 = NULL,
bigM = 100, maxiter = 20, maximize = FALSE)
cc |
defines the linear objective function. |
A |
matrix representing the inequality constraints |
b |
vector, right hand side of the inequalities. |
Aeq |
matrix representing the equality constraints |
beq |
vector, right hand side of the inequalities. |
lb |
lower bounds, if not |
ub |
upper bounds, if not |
x0 |
feasible base vector, will not be used at the moment. |
I0 |
index set of |
bigM |
big-M constant, will be used for finding a base vector. |
maxiter |
maximum number of iterations. |
maximize |
logical; shall the objective be minimized or maximized? |
Solves linear programming problems of the form min cc' * x
such that
A * x \le b
A_{eq} * x = b_{eq}
lb \le x \le ub
List with
x
the solution vector.
fval
the value at the optimal solution.
errno
, mesage
the error number and message.
This is a first version that will be unstable at times. For real linear
programming problems use package lpSolve
.
HwB <hwborchers@googlemail.com>
Vanderbei, R. J. (2001). Linear Programming: Foundations and Extensions. Princeton University Press.
Eiselt, H. A., and C.-L. Sandblom (2012). Operations Research: A Model-based Approach. Springer-Verlag, Berlin Heidelberg.
linprog::solveLP
, lpSolve::lp
## Examples from the book "Operations research - A Model-based Approach"
#-- production planning
cc <- c(5, 3.5, 4.5)
Ain <- matrix(c(3, 5, 4,
6, 1, 3), 2, 3, byrow=TRUE)
bin <- c(540, 480)
linprog(cc, A = Ain, b = bin, maximize = TRUE)
# $x 20 0 120
# $fval 640
#-- diet problem
cc <- c(1.59, 2.19, 2.99)
Ain <- matrix(c(-250, -380, -257,
250, 380, 257,
13, 31, 28), 3, 3, byrow = TRUE)
bin <- c(-1800, 2200, 100)
linprog(cc, A = Ain, b = bin)
#-- employee scheduling
cc <- c(1, 1, 1, 1, 1, 1)
A <- (-1)*matrix(c(1, 0, 0, 0, 0, 1,
1, 1, 0, 0, 0, 0,
0, 1, 1, 0, 0, 0,
0, 0, 1, 1, 0, 0,
0, 0, 0, 1, 1, 0,
0, 0, 0, 0, 1, 1), 6, 6, byrow = TRUE)
b <- -c(17, 9, 19, 12, 5, 8)
linprog(cc, A, b)
#-- inventory models
cc <- c(1, 1.1, 1.2, 1.25, 0.05, 0.15, 0.15)
Aeq <- matrix(c(1, 0, 0, 0, -1, 0, 0,
0, 1, 0, 0, 1, -1, 0,
0, 0, 1, 0, 0, 1, -1,
0, 0, 0, 1, 0, 0, 1), 4, 7, byrow = TRUE)
beq <- c(60, 70, 130, 150)
ub <- c(120, 140, 150, 140, Inf, Inf, Inf)
linprog(cc, Aeq = Aeq, beq = beq, ub = ub)
#-- allocation problem
cc <- c(1, 1, 1, 1, 1)
A <- matrix(c(-5, 0, 0, 0, 0,
0, -4.5, 0, 0, 0,
0, 0, -5.5, 0, 0,
0, 0, 0, -3.5, 0,
0, 0, 0, 0, -5.5,
5, 0, 0, 0, 0,
0, 4.5, 0, 0, 0,
0, 0, 5.5, 0, 0,
0, 0, 0, 3.5, 0,
0, 0, 0, 0, 5.5,
-5, -4.5, -5.5, -3.5, -5.5,
10, 10.0, 10.0, 10.0, 10.0,
0.2, 0.2, 0.2, -1.0, 0.2), 13, 5, byrow = TRUE)
b <- c(-50, -55, -60, -50, -50, rep(100, 5), -5*64, 700, 0)
# linprog(cc, A = A, b = b)
lb <- b[1:5] / diag(A[1:5, ])
ub <- b[6:10] / diag(A[6:10, ])
A1 <- A[11:13, ]
b1 <- b[11:13]
linprog(cc, A1, b1, lb = lb, ub = ub)
#-- transportation problem
cc <- c(1, 7, 4, 2, 3, 5)
Aeq <- matrix(c(1, 1, 1, 0, 0, 0,
0, 0, 0, 1, 1, 1,
1, 0, 0, 1, 0, 0,
0, 1, 0, 0, 1, 0,
0, 0, 1, 0, 0, 1), 5, 6, byrow = TRUE)
beq <- c(30, 20, 15, 25, 10)
linprog(cc, Aeq = Aeq, beq = beq)
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